Question

Find the value of $\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}$.

Answer 1

Hint: First of all convert the whole equation in terms of $\sin \theta$ and $\cos \theta$ by using $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$. Then multiply the numerator and denominator by $2.\dfrac{1}{2}$ and use $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ and $\sin 2A=2\sin A\cos A$ to further solve the question.

Complete step-by-step answer:
Here, we have to find the value of the expression $\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}$. Let us consider the expression given in the question
$E=\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}$

Let us first convert the given expression in terms of $\sin \theta$ and $\cos \theta$. We know that $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$. By using this in the above expression, we get,

& E=\dfrac{\sqrt{3}}{\sin {{20}^{o}}}-\dfrac{1}{\cos {{20}^{o}}} \\\ & E=\dfrac{\left( \sqrt{3}\cos {{20}^{o}}-\sin {{20}^{o}} \right)}{\left( \sin {{20}^{o}} \right).\left( \cos {{20}^{o}} \right)} \\\ \end{aligned}$$ By multiplying $$2\times \dfrac{1}{2}$$ on both numerator and denominator of the above expression, we get, $$E=\dfrac{2\times \dfrac{1}{2}\left( \sqrt{3}\cos {{20}^{o}}-\sin {{20}^{o}} \right)}{2\times \dfrac{1}{2}\left( \sin {{20}^{o}} \right).\left( \cos {{20}^{o}} \right)}$$ We can also write the above expression as, $$E=\dfrac{2\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{o}}-\dfrac{1}{2}\sin {{20}^{o}} \right)}{\dfrac{1}{2}\left( 2\sin {{20}^{o}}.\cos {{20}^{o}} \right)}$$ $$\Rightarrow E=\dfrac{4\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{o}}-\dfrac{1}{2}\sin {{20}^{o}} \right)}{\left( 2\sin {{20}^{o}}.\cos {{20}^{o}} \right)}$$ We know that $$\sin {{60}^{o}}=\dfrac{\sqrt{3}}{2}$$ and $$\cos {{60}^{o}}=\dfrac{1}{2}$$. By using this in the above expression, we get, $$E=\dfrac{4\left( \sin {{60}^{o}}\cos {{20}^{o}}-\cos {{60}^{o}}\sin {{20}^{o}} \right)}{\left( 2\sin {{20}^{o}}.\cos {{20}^{o}} \right)}$$ We know that $$\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$$. By using it in the numerator of the above expression, we get, $$E=\dfrac{4\left( \sin \left( {{60}^{o}}-{{20}^{o}} \right) \right)}{2\sin {{20}^{o}}\cos {{20}^{o}}}$$ So, we get, $$E=\dfrac{4\left( \sin {{40}^{o}} \right)}{2\sin {{20}^{o}}\cos {{20}^{o}}}$$ Now, we know that $$2\sin A\cos A=\sin 2A$$. By using this in the denominator of the above expression, we get, $$E=\dfrac{4\left( \sin {{40}^{o}} \right)}{\sin \left( 2\left( {{20}^{o}} \right) \right)}$$ $$\Rightarrow E=\dfrac{4\left( \sin {{40}^{o}} \right)}{\left( \sin {{40}^{o}} \right)}$$ By canceling the like terms from the above expression, we get E = 4. So, we get the value of $$\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}=4$$. Note: Students must remember the general trigonometric formulas like $$\sin \left( A\pm B \right)=\sin A\cos B\pm \cos A\sin B$$ and $$\sin 2A=2\sin A\cos A$$. Also, students should keep in mind the values of sine and cosine at general angles like $${{0}^{o}},{{30}^{o}},{{60}^{o}},{{45}^{o}}\text{ and }{{90}^{o}}$$. Always try to convert the given expression into some trigonometric formulas in these types of questions.